In this chapter we show how a number of heat conduction problems are solved by an analogous procedure: ◦ (i) an energy balance is made over a thin slab or shell perpendicular to the direction of the heat flow, and this balance leads to a first-order differential equation from which the heat flux distribution is obtained ◦ (ii) then into this expression for the heat flux, we substitute Fourier's law of heat conduction, which gives a first-order differential equation for the temperature as a function of position
The integration constants are then determined by use of boundary conditions for the temperature or heat flux at the bounding surfaces.
We select a slab (or shell), the surfaces of which are normal to the direction of heat conduction, then for steady-state (i.e., time-independent) systems:
These three terms can be added to give the "combined energy flux" e. Note that in non-flow systems (for which v is zero) the e vector simplifies to the q vector, which is given by Fourier's law. The energy production term in Eq. 10.1-1 includes: ◦ (i) the degradation of electrical energy into heat ◦ (ii) the heat produced by slowing down of neutrons and nuclear fragments liberated in the fission process ◦ (iii) the heat produced by viscous dissipation ◦ (iv) the heat produced in chemical reactions
After Eq. 10.1-1 has been written for a thin slab or shell of material, the thickness of the slab or shell is allowed to approach zero. This procedure leads ultimately to an expression for the temperature distribution containing constants of integration, which we evaluate by use of boundary conditions.
The commonest types of boundary conditions are: ◦ The temperature may be specified at a surface ◦ The heat flux normal to a surface may be given (this is equivalent to specifying the normal component of the temperature gradient ◦ At interfaces the continuity of temperature and of the heat flux normal to the interface are required ◦ At a solid-fluid interface, the normal heat flux component may be related to the difference between the solid surface temperature, To and the "bulk" fluid temperature, Tb Newton's law of cooling
The first system we consider is an electric wire of circular cross section with radius R and electrical conductivity k, ohm-1 cm-1. Through this wire there is an electric current with current density, I, amp/cm2.
The transmission of an electric current is an irreversible process, and some electrical energy is converted into heat (thermal energy).
The rate of heat production per unit volume is given by the expression: resulting from electrical dissipation
Assume that the temperature rise in the wire is not so large that the temperature dependence of either the thermal or electrical conductivity need be considered. The surface of the wire is maintained at temperature To. How to find the radial temperature distribution within the wire?
For the energy balance we take the system to be a cylindrical shell of thickness ∆r and length L. Since v = 0 in this system, the only contributions to the energy balance is q. why?
We now substitute these quantities into the energy balance.
Division by 2 L∆r and taking the limit as ∆r goes to zero gives:
Then, we get:
Integrated to give:
The integration constant C, must be zero because of the boundary condition that:
Then, we get:
This states that the heat flux increases linearly with r.
We now substitute Fourier's law in the form qr = -k(dT/dr),
Integrated to give:
Finally we get:
From temperature profile, we can obtained:
There is, after all, a pronounced similarity between the heated wire problem and the viscous flow in a circular tube. Only the notation is different:
There are many examples of heat conduction problems in the electrical industry. The minimizing of temperature rises inside electrical machinery prolongs insulation life.
A copper wire has a radius of 2 mm and a length of 5 m. For what voltage drop would the temperature rise at the wire axis be 10°C, if the surface temperature of the wire is 20°C? Given
Repeat the analysis in section 10.2, assuming that To is not known, but that instead the heat flux at the wall is given by Newton's "law of cooling" (Eq. 10.1-2). Assume that the heat transfer coefficient h and the ambient air temperature Tair are known.
We consider the flow of an incompressible Newtonian fluid between two coaxial cylinders as shown in Fig. 10.4-1. The surfaces of the inner and outer cylinders are maintained at T = To and T = Tb, respectively. We can expect that T will be a function of r alone.
Modification of a portion of the flow system in Fig. 10.41, in which the curvature of the bounding surfaces is neglected.
As the outer cylinder rotates, each cylindrical shell of fluid "rubs" against an adjacent shell of fluid This friction between adjacent layers of the fluid produces heat; that is, the mechanical energy is degraded into thermal energy The volume heat source resulting from this "viscous dissipation," which can be designated by Sv appears automatically in the shell balance when we use the combined energy flux vector e defined at the
If the slit width b is small with respect to the radius R of the outer cylinder, then the problem can be solved approximately by using the somewhat simplified system depicted in Fig. 10.42. That is, we ignore curvature effects and solve the problem in Cartesian coordinates. The velocity distribution is then vz = vb(x/b), where vb = ΩR. We now make an energy balance over a shell of thickness ∆x, width W, and length L. Since the fluid is in motion, we use the combined energy flux vector e
Dividing by WL ∆x and letting the shell thickness ∆x go to zero then gives:
Integration:
